We can do a little better. Recall that $\varphi(n)$ counts the integers from $0$ to $n-1$ that are relatively prime to $n$. Note also that if $e$ is positive, then $x$ and $m$ are relatively prime iff $x$ and $m^e$ are relatively prime. This is because $x$ and $m$ are relatively prime iff they have no common prime factor. But that is the case iff $x$ and $m^e$ have no common prime factor.

Let $0 \le x < m$, where $m>1$. Then $x$ is relatively prime to $m$ iff $x+m$ is relatively prime to $m$ iff $x+2m$ is relatively prime to $m$, and so on. It follows that for any $a$, the number of integers in the chunk $[a,a+m-1]$ that are relatively prime to $m$ is the same as the number of integers in the chunk $[a+m,a+2m-1]$ that are relatively prime to $m$.

Divide the interval from $0$ to $m^p-1$ into $m^{p-1}$ equal-sized chunks. Each chunk has $\varphi(m)$ numbers relatively prime to $m$. So up to $m^p-1$ there are $\varphi(m)\,m^{p-1}$ numbers relatively prime to $m$. It follows that
$$\varphi(m^p)=\varphi(m)\,m^{p-1}.$$
Similarly, $\varphi(m^q)=\varphi(m)\,m^{q-1}$. It follows that
$$\frac{\varphi(m^q)}{\varphi(m^p)}=m^{q-p}.$$